Theoretical Computer Science 349 (2005) 382 – 391
www.elsevier.com/locate/tcs
The complexity of Kemeny elections夡
Edith Hemaspaandraa , Holger Spakowskib,∗,1 , Jörg Vogelc
a Department of Computer Science, Rochester Institute of Technology, Rochester, NY 14623, USA
b Institut für Informatik, Heinrich-Heine-Universität Düsseldorf, 40225 Düsseldorf, Germany
c Institut für Informatik, Friedrich-Schiller-Universität Jena, 07740 Jena, Germany
Received 12 December 2003; accepted 29 August 2005
Communicated by B. Rovan
Abstract
Kemeny proposed a voting scheme which is distinguished by the fact that it is the unique voting scheme that is neutral, consistent,
and Condorcet. Bartholdi, Tovey, and Trick showed that determining the winner in Kemeny’s system is NP-hard. We provide a
stronger lower bound and an upper bound matching the lower bound, namely, we show that determining the winner in Kemeny’s
system is complete for PNP , the class of sets solvable via parallel access to NP.
© 2005 Elsevier B.V. All rights reserved.
Keywords: Computational complexity; Election systems; Parallel access to NP
1. Introduction
We investigate the complexity of determining the winner of Kemeny’s voting scheme. Kemeny’s voting system
is distinguished by the fact that it is the unique voting system which is neutral, consistent, and Condorcet. With
voting scheme here is meant a rule for choosing winning candidates (or alternatives) based on the voter preference
rankings.
For the case of more than two candidates, it is not trivial to say what the right way of evaluating the rankings of the
voters is. As an example of the difficulties occurring, see the prominent Condorcet Paradox [3] dating back to 1785:
assume there are 3n voters with the preference rankings a > b > c, b > c > a, and c > a > b, each given by exactly
n voters. If we evaluate the rankings by majority rule, then we get a cyclic aggregate preference ranking: a defeats b,
b defeats c, and c defeats a.
Social choice theorists have investigated various different voting schemes, each having its specific advantages
and disadvantages. There are a number of “reasonable” properties that one would like a voting system to satisfy,
夡
Research supported in part by Grants NSF-CCR-0311021, NSF-INT-9815095/DAAD-315-PPP-gü-ab, a grant from the DAAD, DFG project
RO 1202/9-1, and an RIT FEAD grant. Some of these results were presented at MFCS 2000 and WAIT 2001.
∗ Corresponding author.
E-mail addresses: [email protected] (E. Hemaspaandra), [email protected] (H. Spakowski), [email protected] (J. Vogel).
1 Work done in part while visiting the University of Rochester.
0304-3975/$ - see front matter © 2005 Elsevier B.V. All rights reserved.
doi:10.1016/j.tcs.2005.08.031
E. Hemaspaandra et al. / Theoretical Computer Science 349 (2005) 382 – 391
383
including nondictatorship, monotonicity, independence of irrelevant alternatives, and the Pareto principle. Arrow’s
famous Impossibility Theorem [1] states that there is no voting scheme enjoying all the just mentioned properties.
The investigation of computational complexity issues involved in voting schemes is a relatively young field. Automated group decision making is becoming an important issue, which calls for an investigation of voting rules from a
computational complexity point of view. Bartholdi et al. [2] studied a voting system suggested by Dodgson (who is
better known by his pen name, Lewis Carroll) in 1876. In that paper, it was shown that it is NP-hard to determine whether
a given candidate has won a given election in Lewis Carroll’s voting scheme. Hemaspaandra et al. [6] improved that
result by raising the lower bound to PNP . They also proved a matching upper bound, thus obtaining PNP -completeness
for Dodgson’s voting scheme.
Bartholdi et al. [2] also investigated Kemeny’s voting scheme. Kemeny’s voting scheme, which will be described
in Section 2, was introduced by Kemeny [9] and specified by Levenglick [11]. Young and Levenglick [18] showed
that Kemeny’s voting scheme is the unique voting scheme that is neutral, consistent, and Condorcet. Bartholdi et al.
showed that determining the winner in Kemeny’s voting scheme is NP-hard. The exact complexity of Kemeny’s voting
scheme however remained an open problem. We show that the winner problem for Kemeny’s voting scheme is PNP complete.
Another voting system whose computational complexity has recently been investigated is Young’s voting scheme
[14]. Young’s voting scheme also turned out to be PNP -complete. This raises the question of what is special about PNP
that all these voting schemes are complete for this class. The above-mentioned voting schemes have in common that
they assign a hard-to-compute score to each candidate, and the candidates with lowest (or highest in case of Young’s
voting system) score are the winners. The value of the score is in all three cases polynomially bounded in the size of
the input. That easily yields PNP as an upper bound. It is known that “comparison versions” of NP-hard optimization
problems often are complete for PNP (for instance, the problem of comparing the maximum vertex cover sizes of two
given graphs). That gives some hint that these voting schemes might be PNP -complete. However, the actual proofs of
the PNP lower bounds are complicated and are substantially different in each case. In particular, these problems do not
easily reduce to each other.
This paper is organized as follows. In Section 2, we define Kemeny’s voting scheme and the decision problems
that we study. In Section 3, we give some background on the class PNP and outline the proof of the main result. In
Section 4, we prove our main result.
2. Kemeny’s voting scheme
Like Dodgson’s and Young’s voting schemes, Kemeny’s voting scheme is a preferential voting system. Each voter
casts his or her vote by ranking all the candidates in order of preference. Ties are explicitly allowed. 2 For example, a
voter may rank candidates a, b, c, d, and e by the preference ranking a > b = c > d > e. Candidate a is the favorite,
and e is the least favorite candidate in this ranking. Candidates b and c are considered to be of equal desirability, i.e.,
they are tied. A preference ranking without ties is called a strict preference ranking. We identify each voter with its
preference ranking, and we will view the set of voters as a multiset of preference rankings.
Kemeny defined the outcome of an election as the collection of preference rankings that are “closest” to the preference
rankings of the voters. Such a preference ranking is called a Kemeny consensus. A candidate is a winner of the election
if it is a preferred candidate in a Kemeny consensus.
There are different ways to define closeness. For Kemeny elections, the goal is to minimize the Kemeny score: the
sum of the distances to the preference rankings of the voters.
For each pair P, Q of preference rankings we define the distance
dist(P , Q) =
dP ,Q (c, d),
{c,d}
2 This follows Kemeny’s original definition [9]. It should be noted that there is no consensus about this in the literature. To avoid confusion, we
will use the term preference ranking rather than preference order. It follows immediately from our proofs that our complexity results go through if
we do not allow ties in preference rankings.
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E. Hemaspaandra et al. / Theoretical Computer Science 349 (2005) 382 – 391
where the sum is taken over all unordered pairs {c, d} of candidates, and
⎧
0 if P and Q agree on c and d,
⎪
⎪
⎨
1 if one of P or Q has a preference among c and d and the
dP ,Q (c, d) =
other has not,
⎪
⎪
⎩
2 if P andQ strictly disagree on c and d.
Given a set of candidates C and a multiset of preference rankings V on C, we define the following three Kemeny
score functions:
• For every preference ranking P on C,
KemenyScore(C, V , P ) =
dist(P , Q).
Q∈V
• For every candidate c ∈ C,
KemenyScore(C, V , c) = min{KemenyScore(C, V , P ) | P is a preference
ranking on C, and c is a preferred candidate in P}.
•
KemenyScore(C, V ) = min{KemenyScore(C, P , V ) | P is a preference ranking on C }.
We define the following decision problems related to Kemeny elections:
Decision problem: Kemeny Score
Instance: A set of candidates C; a multiset V of preference rankings on C; a positive integer k.
Question: Is KemenyScore(C, V )k?
Decision problem: Candidate Kemeny Score
Instance: A set of candidates C; a multiset V of preference rankings on C; a candidate c ∈ C; a positive integer k.
Question: Is KemenyScore(C, V , c)k?
Decision problem: Kemeny Winner
Instance: A set of candidates C; a multiset V of preference rankings on C; a candidate c ∈ C.
Question: Is there some Kemeny consensus P in which no candidate is strictly preferred to c? Equivalently, is
KemenyScore(C, V , c)KemenyScore(C, V , d) for all d ∈ C?
Decision problem: Kemeny Ranking
Instance: A set of candidates C; a multiset V of preference rankings on C; two distinguished candidates c, d ∈ C.
Question: Does c tie-or-defeat d in the election? That is, is KemenyScore(C, V , c) KemenyScore(C, V , d)?
As is usual, we represent the multisets of preference rankings by lists of preference rankings. Multiplicities of the
same preference ranking are given by repetition (i.e, not by a number in binary). This representation corresponds to
elections in which traditional paper ballots are used. In such elections, the input size is naturally given by the number
of candidates multiplied by the number of voters. 3
Bartholdi et al. [2] showed that Kemeny Score is NP-complete and that Kemeny Winner and Kemeny
Ranking are NP-hard. We now state the main result of this paper.
Theorem 2.1. Kemeny Winner and Kemeny Ranking are PNP -complete.
Corollary 2.2. If Kemeny Winner is NP-complete, then PH = NP.
3 The issue of studying the complexity of voting systems when the representation is succinct was raised in a technical report precursor of [6] for
the case of Dodgson elections and is also interesting in the current case of Kemeny elections. As pointed out by an anonymous referee, when the
multiplicities in the multisets are given in binary, Kemeny Winner and Kemeny Ranking are in PNP . Clearly, our PNP lower bounds transfer
to this case, but the exact complexity of these problems remains an open question.
E. Hemaspaandra et al. / Theoretical Computer Science 349 (2005) 382 – 391
385
3. Parallel access to NP
PNP is the class of sets solvable by some P oracle machine that, instead of asking its oracle queries sequentially,
accesses its NP oracle in parallel. This type of access is also known as truth-table access, which was introduced by
Ladner et al. [10], and the class is also written as PttNP .
Clearly, NP ∪ coNP ⊆ PNP ⊆ PNP , and all these inclusions are believed to be strict. PNP is an extremely robust
class, that has many different characterizations. For example, Hemaspaandra [7] proved that PNP is equal to the class
PNP[log] (the class of languages that can be solved via O(log n) queries to an NP oracle), which was introduced by
p
p
Papadimitriou and Zachos [13]. Wagner [17] introduced the name 2 for PNP[log] and proved that 2 = LNP .
NP
Wagner provided a useful tool to prove P -hardness and applied it to prove dozens of problems complete for PNP
[16], though none of these problems were particularly natural. Hemaspaandra et al. used Wagner’s tool to prove PNP completeness for Lewis Caroll’s voting scheme [6]. Very recently [14] stated PNP -completeness for Young’s voting
scheme. There also exist natural PNP -complete problems in artificial intelligence and modal logic (see [4]).
In [2], NP-hardness for Kemeny Score is proved by a reduction from the NP-complete digraph problem Feedback
Arc Set. Our approach to prove PNP -hardness for Kemeny Winner is the following. Define a PNP -complete version Feedback Arc Set Member of Feedback Arc Set that is PNP -complete and adapt the reduction from
Feedback Arc Set to Kemeny Score so that it becomes a reduction from Feedback Arc Set Member
to Kemeny Winner and from Feedback Arc Set Member to Kemeny Ranking.
There exists a long list of NP-complete problems, but there are not that many known PNP -complete problems. In
particular, no PNP -complete version of Feedback Arc Set was previously known. Fortunately, there are known
PNP -complete vertex cover problems, and we are able to turn Karp’s [8] reduction from Vertex Cover to Feedback
Arc Set into a reduction from a PNP -complete vertex cover problem to Feedback Arc Set Member.
4. Kemeny Winner and Kemeny Ranking are complete for PNP
In this section, we will prove the main result of this paper, namely that Kemeny Winner and Kemeny Ranking
are complete for PNP . It is easy to show that Kemeny Winner and Kemeny Ranking are in PNP . The PNP algorithms will use the set Candidate Kemeny Score as an oracle. Note that Candidate Kemeny Score
is clearly in NP. Let C be a set of candidates and V a multiset of preference rankings on C. For every c ∈ C,
KemenyScore(C, V , c) V ·C2 , so we can in polynomial time in parallel query all tuples C, V , c, k to Candidate
Kemeny Score for all c ∈ C and all k V · C2 . With the answers to all these queries in hand, we know
KemenyScore(C,V ,c) for each candidate c∈C, since this is the smallest k such that C,V ,c,k∈Candidate Kemeny
Score. From the Kemeny scores for the candidates, it is trivial to verify that a certain candidate is a winner (has the
smallest Kemeny score) or that a certain candidate ties-or-defeats another (does not have a higher Kemeny score).
It remains to show that Kemeny Winner and Kemeny Ranking are hard for PNP . The hardness proof consists
of two parts. In Section 4.1 we give a reduction from the problem Feedback Arc Set Member to Kemeny
Ranking and Kemeny Winner. The PNP -hardness of Feedback Arc Set Member is shown in Section 4.2.
p
4.1. Feedback Arc Set Member m Kemeny Winner and Feedback Arc Set Member
p
m Kemeny Ranking
In [2], NP-hardness for Kemeny Score is proved by a reduction from the NP-complete digraph problem Feedback
Arc Set, which will be defined below. Our approach to prove PNP -hardness for Kemeny Winner is the following.
Define a version Feedback Arc Set Member of Feedback Arc Set that is PNP -complete and adapt the
reduction from Feedback Arc Set to Kemeny Score so that it becomes a reduction from Feedback Arc
Set Member to Kemeny Winner. We will then use this reduction to obtain a reduction from Feedback Arc
Set Member to Kemeny Ranking.
We will start by defining Feedback Arc Set and Feedback Arc Set Member.
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E. Hemaspaandra et al. / Theoretical Computer Science 349 (2005) 382 – 391
Definition 4.1. (1) A feedback arc set (FAS) for a digraph G is a set of arcs of G that includes at least one arc from
every cycle in G.
(2) Feedback Arc Set = {G, k | G is a digraph, k a positive integer, and G has a feedback arc set of size at
most k}.
(3) Feedback Arc Set Member = {G, v | G is an irreflexive and antisymmetric digraph, v is a vertex of G,
and some minimum size feedback arc set of G contains all arcs entering v}.
p
We will now prove that Feedback Arc Set Member m Kemeny Winner.
p
Lemma 4.2. Feedback Arc Set Member m Kemeny Winner.
Proof. We will modify the reduction from Feedback Arc Set to Kemeny Score from [2] to construct a reduction f from Feedback Arc Set Member to Kemeny Winner. In our reduction, the vertices of the graph will
correspond to the candidates in the election. Suppose that we are given G, ĉ, where G = C, A is an irreflexive and
antisymmetric digraph, and ĉ ∈ C.
Due to a note by McGarvey [12], we can interpret G as an election. We can in polynomial time compute an election
g(G) = C, V, where V is a multiset of strict preference rankings on C such that the number of voters is even and
[c → d] ∈ A iff V /2 + 1 voters prefer c to d (and V /2 − 1 voters prefer d to c), and c and d are unconnected
iff exactly half of the voters prefer c to d (and exactly half of the voters prefer d to c): for each arc [1 → 2 ] ∈ A
we create one voter with preference ranking 1 > 2 > c1 > c2 > · · · > cn−2 and one voter with preference ranking
cn−2 > · · · > c2 > c1 > 1 > 2 , where c1 , . . . , cn−2 are the remaining candidates.
Let f (G, ĉ) = g(G), ĉ = C, V , ĉ. We have to show that G, ĉ ∈ Feedback Arc Set Member if and only
if C, V , ĉ ∈ Kemeny Winner.
Lemma 3 in [2] can easily be strengthened such that the particular Kemeny winner is preserved. We get the following
lemma.
Lemma 4.3 (Bartholdi et al. [2]). Given a set of candidates C and a multiset V of strict preference rankings on C. If
c ∈ C is a Kemeny winner, then there exists a strict Kemeny consensus 4 such that c is the preferred candidate in the
consensus. Our election C, V consists only of strict preference rankings. That justifies the definition of the following score
function.
• StrictKemenyScore(C, V , c) = min{KemenyScore(C, V , P ) | P is a strict preference ranking on C, and c is the
preferred candidate in P }.
Lemma 4.3 implies that the winners of C, V are the candidates with smallest StrictKemenyScore, because
StrictKemenyScore(C, V , c) = KemenyScore(C, V , c) for all Kemeny winners c.
We define the following functions in analogy to KemenyScore.
• For every strict preference ranking P on C,
Disagree(G, P ) = {[c → d] | [c → d] ∈ A and P prefers d to c}.
• For every candidate c ∈ C,
Disagree(G, c) = min{Disagree(G, P ) | P is a strict preference ranking on C,
and c is the preferred candidate in P }.
The following claim is implicit in [2].
Claim 4.4. Let g(G) = C, V. Then the following hold.
(1) For each strict preference ranking P,
KemenyScore(C, V , P ) = FixedCost(G) + 4 Disagree(G, P ).
4 A strict Kemeny consensus is a Kemeny consensus that is a strict preference ranking.
E. Hemaspaandra et al. / Theoretical Computer Science 349 (2005) 382 – 391
387
(2) For any candidate c ∈ C,
StrictKemenyScore(C, V , c) = FixedCost(G) + 4 Disagree(G, c),
where FixedCost is a function that depends neither on P nor on c.
Proof. (1) According to the definitions,
KemenyScore(C, V , P ) =
dist(P , Q) =
dP ,Q (c, d).
Q∈V
{c,d} Q∈V
Given an unordered pair {c, d}, it is easy to see that for every strict preference ranking P,
⎧
V /2 − 1 if [c → d] ∈ A and P prefers c to d,
⎪
⎪
⎪
⎪
V
/2 + 1 if [c → d] ∈ A and P prefers d to c,
⎨
dP ,Q (c, d) = 2 · V /2 − 1 if [d → c] ∈ A and P prefers d to c,
⎪
⎪
V /2 + 1 if [d → c] ∈ A and P prefers c to d,
Q∈V
⎪
⎪
⎩
V /2
if [c → d] ∈
/ A and [d → c] ∈
/ A.
Hence
KemenyScore(C, V , P )
= 2 · (V /2 · {{x, y} | x = y, [x → y] ∈
/ A and [y → x] ∈
/ A} + (V /2 − 1) A + 2 Disagree(G, P ))
= FixedCost(G) + 4 Disagree(G, P ).
Clearly, FixedCost(G) does not depend on P.
(2) By definition,
StrictKemenyScore(C, V , c)
= min{KemenyScore(C, V , P ) | P is a strict preference ranking on C,
and c is the preferred candidate in P }
= min{FixedCost(G) + 4 Disagree(G, P ) | P is a strict preference
ranking on C, and c is the preferred candidate in P },
due to Claim 4.4(1). By the definition of Disagree(G, c), it follows that
StrictKemenyScore(C, V , c) = FixedCost(G) + 4 Disagree(G, c).
Hence Claim 4.4 is proved.
Claim 4.5. Disagree(G, c)= min {F | F is a FAS of G containing all arcs entering c} .
Proof. To prove that the left-hand side is less than or equal to the right-hand side, suppose that F is a FAS of G that
be the digraph that is obtained from G if we throw away all arcs that belong to
contains all arcs entering c. Let G
does not contain cycles and G
does not contain any arcs entering c. Order C, the set of vertices of G,
as
F. Then G
i < j . This is possible because G
is
c1 , c2 , . . . , c , where = C, such that c1 = c and for all arcs [ci → cj ] in G,
Now consider
cycle-free; therefore, its transitive closure is a partial order, any total extension of which agrees with G.
P ) = 0. Since G consists of G
the preference ranking P = c1 > c2 > · · · > c−2 > c−1 > c . Clearly, Disagree(G,
plus F extra arcs, it follows that Disagree(G, P ) F , and therefore Disagree(G, c)F .
= c1 > c2 > · · · > c−2 >
To prove that the left-hand side is greater than or equal to the right-hand side, let P
be the graph we
c−1 > c be a preference ranking on C with c1 = c and Disagree(G, P ) = Disagree(G, c). Let G
obtain if we delete from G the Disagree(G, c) arcs in F = {[cj → ci ] | [cj → ci ] ∈ A and i < j } (the arcs disagreeing
). Graph G
does not contain cycles, is obtained from G by removing Disagree(G, c) arcs, and does not contain
with P
). The arc set F
is a FAS for G. Moreover, F
contains
any arcs entering c (since arcs entering c would disagree with P
all arcs entering c, because G with the arcs in F removed has no arcs entering c. 388
E. Hemaspaandra et al. / Theoretical Computer Science 349 (2005) 382 – 391
Using Lemma 4.3 and Claims 4.4 and 4.5, it is easy to prove that the following statements are equivalent:
(i) Candidate ĉ is a Kemeny winner of the election C, V.
(ii) Candidate ĉ has smallest KemenyScore(C, V , ĉ) (from the definition of Kemeny winner).
(iii) Candidate ĉ has smallest StrictKemenyScore(C, V , ĉ) (Lemma 4.3).
(iv) Candidate ĉ has smallest Disagree(G, ĉ) (Claim 4.4(2)).
(v) Candidate ĉ has smallest
min{F | F is a FAS of G containing all arcs entering ĉ}
(Claim 4.5).
(vi) There is a minimum size FAS of G containing all arcs entering ĉ.
To see (v) → (vi), note that for every FAS of G there exists a vertex v such that F contains all arcs entering v. That
concludes the proof of Lemma 4.2. 5 We can use the reduction of the previous lemma to obtain a reduction from Feedback Arc Set Member to
Kemeny Ranking. The main idea is to add a special candidate d̂ that is always a winner of the election. This way,
ĉ is a winner of the election if and only if ĉ ’s KemenyScore is not greater than d̂ ’s KemenyScore.
p
Lemma 4.6. Feedback Arc Set Member m Kemeny Ranking.
Proof. Suppose that x = G, ĉ, with G an irreflexive, antisymmetric digraph and ĉ ∈ V (G). It is easy to see that the
following hold:
(1) G has a minimum size feedback arc set that contains all vertices entering ĉ if and only if (G ∪ {d̂}, ∅) has a
minimum size feedback arc set that contains all vertices entering ĉ. 6
(2) Any minimum size feedback arc set of (G∪{d̂}, ∅) contains all arcs entering d̂ (since there are no arcs entering d̂).
Let C, V = g(G ∪ {d̂}, ∅) where g is defined as in the proof of Lemma 4.2. Then C, V , ĉ = f (G ∪ {d̂}, ∅, ĉ) and
C, V , d̂ = f (G ∪ {d̂}, ∅, d̂), where f is the reduction from Feedback Arc Set Member to Kemeny Winner
from Lemma 4.2. From the observations above, and the fact that f is a reduction from Feedback Arc Set Member
to Kemeny Winner, it follows that:
(1) G has a minimum size feedback arc set that contains all vertices entering ĉ if and only if ĉ is a winner of C, V.
(2) d̂ is a winner of C, V.
It follows immediately that G has a minimum size feedback arc set that contains all vertices entering ĉ if and only if
KemenyScore(C, V , ĉ) KemenyScore(C, V , d̂).
4.2. Feedback Arc Set Member is PNP -hard
In order to conclude from Lemmas 4.2 and 4.6 that Kemeny Winner and Kemeny Ranking are PNP -hard, we
still need to show that Feedback Arc Set Member is PNP -hard. Karp [8] proved that Feedback Arc Set is
NP-hard by reducing Vertex Cover to it. We will follow the same approach as in Section 4.1: we will define a PNP complete version Vertex Cover Member of Vertex Cover, and we will reduce Vertex Cover Member
to Feedback Arc Set Member. Here are the definitions:
Definition 4.7. (1) A vertex cover of graph G is a subset of vertices that contains at least one endpoint of every edge
of G.
(2) Vertex Cover = {G, k | G is a graph, k a positive integer, and G has a vertex cover of size at most k}.
(3) Vertex Cover Member = {G, v | G is a graph, v is a vertex of G, and some minimum size vertex cover of
G contains v}.
5 At this point, it is easy to see that Kemeny’s voting scheme is also PNP -complete if we disallow ties in all preference rankings (cf. footnote 2).
6 We use the standard definition of ∪ on graphs: for two disjoint graphs G
and H
, V (G
∪H
) = V (G)
∪ V (H
), and E(G
∪H
) = E(G)
∪ E(H
).
E. Hemaspaandra et al. / Theoretical Computer Science 349 (2005) 382 – 391
389
p
Lemma 4.8. Vertex Cover Member m Feedback Arc Set Member.
Proof. We will use Karp’s reduction from Vertex Cover to Feedback Arc Set [8]. Given an (undirected)
graph G, define digraph H = W, A as follows.
• W = {v, v | v ∈ V (G)}, and
• A = {[v → v ] | v ∈ V (G)} ∪ {[v → w], [w → v] | {v, w} ∈ E(G)},
where v is a duplicate of v for each vertex v. From [8], we know that G contains a vertex cover of size at most k if and
only if H contains a feedback arc set of size at most k. Note that this implies that the minimum size of a vertex cover
for G is the same as the minimum size of a feedback arc set for H.
We modify Karp’s reduction to obtain a reduction from Vertex Cover Member to Feedback Arc Set
Member as follows. For G a graph, and v̂ ∈ V (G), let f (G, v̂) = H, v̂ , where H is the digraph from Karp’s
reduction defined above. Note that H is irreflexive and antisymmetric and that f is computable in polynomial time.
It remains to show that G, v̂ ∈ Vertex Cover Member if and only if H, v̂ ∈ Feedback Arc Set
Member. This follows almost immediately from the following lemma.
Lemma 4.9. G has a vertex cover of size at most k containing v̂ if and only if H has a feedback arc set of size at most
k containing [v̂ → v̂ ].
This completes the proof of Lemma 4.8, since G, v̂ ∈ Vertex Cover Member if and only if G has a
minimum size vertex cover containing v̂. By Lemma 4.9 and the properties of Karp’s reduction, this holds if and
only if H has a minimum size feedback arc set containing [v̂ → v̂ ]. This last step is equivalent to H, v̂ ∈
Feedback Arc Set Member, since [v̂ → v̂ ] is the only arc entering v̂ . Proof of Lemma 4.9. The proof follows from inspection of Karp’s proof [8]. We include the proof for the sake of
completeness. First suppose that V ⊆ V (G) is a vertex cover for G such that V = k and v̂ ∈ V . We claim that
{[v → v ] | v ∈ V } is a feedback arc set for H. Note that the size of this set is k, and that [v̂ → v̂ ] is in this set.
Suppose for a contradiction that W, A \ {[v → v ] | v ∈ V } contains a cycle. By construction of H, this cycle is of
the following form for some k 2, vi = vj for all i = j , and all vi not in V :
v1 → v1 → v2 → v2 → · · · → vk → vk → v1 .
But in G, this implies that {v1 , v2 } ∈ E, while v1 , v2 ∈ V . This contradicts the fact that V is a vertex cover for G.
For the converse, suppose that H has a feedback arc set A of size k that contains [v̂ → v̂ ]. We claim that
V = {v ∈ V | ∃w ∈ W : [v → w] ∈ A or [v → w] ∈ A }
forms a vertex cover. Note that V k and that v̂ ∈ V .
Suppose for a contradiction that {v1 , v2 } ∈ E and v1 , v2 ∈ V . Then (by definition)
[v1 → v1 ], [v1 → v2 ], [v2 → v2 ], [v2 → v1 ] ∈ A \ A
so that we have a cycle in W, A \ A, which contradicts our assumption that A is a feedback arc set.
It remains to show that Vertex Cover Member is PNP -hard. In general, there is often a “comparison version”
of an NP-complete problem that is PNP -complete. Wagner [16] stated that problems like {G, H | G and H are graphs
such that the sizes of G’s and H’s minimum vertex covers are the same} are PNP -complete. 7 A full proof of the PNP completeness of the closely related problem of determining whether two graphs have maximum independent sets of the
same size can be found in Appendix A of [7]. Spakowski and Vogel showed that the following problem is PNP -complete
as well [15].
Definition 4.10. Min Card Vertex Cover Compare = {G, H | G and H are graphs such that the size of G’s
minimum vertex cover is less than or equal to the size of H’s minimum vertex cover}.
7 Wagner states completeness for PNP , a class that was later shown to be equivalent to PNP [17].
bf
390
E. Hemaspaandra et al. / Theoretical Computer Science 349 (2005) 382 – 391
Lemma 4.11 (Spakowski and Vogel [15]). Min Card Vertex Cover Compare is PNP -complete. The problem
remains PNP -complete if we only allow pairs of graphs G, H such that V (G) = V (H ).
p
Lemma 4.12. Min Card Vertex Cover Compare m Vertex Cover Member.
Proof. Let G and H be graphs such that V (H ) = V (G). Let v and w be two new vertices and let F = (G ∪
{v}, ∅) ⊕ (H ∪ {w}, ∅). 8 We claim that the following holds: G, H ∈ Min Card Vertex Cover Compare
if and only if F has a minimum vertex cover that contains w. Our reduction will map G, H to F, w.
a graph, let (G)
be the minimum size of a vertex cover for G.
Note that V is a vertex cover of F if and only
For G
if one of the two following statements holds:
(1) V contains all vertices of (G ∪ {v}, ∅) and V contains a vertex cover of (H ∪ {w}, ∅). The smallest vertex cover
of this form is of size V (G) + 1 + (H ∪ {w}, ∅) = V (H ) + 1 + (H ). Note that w is not part of the
smallest vertex cover of this type.
(2) V contains all vertices of (H ∪ {w}, ∅) and V contains and a vertex cover of (G ∪ {v}, ∅). The smallest vertex
cover of this form is of size V (H ) + 1 + (G ∪ {v}, ∅) = V (H ) + 1 + (G). Note that w is always part of
a vertex cover of this type.
It follows that w is an element of a minimum size vertex cover of F if and only if there is a minimum size vertex cover
of type 2 described above. This is the case if and only if (G)(H ). Lemmas 4.11, 4.12, 4.8, 4.2, and 4.6 immediately imply that Kemeny Winner and Kemeny Ranking are PNP hard. This completes the proof of Theorem 2.1. In addition, these lemmas and the upper bounds for Kemeny Winner
and Kemeny Ranking proven at the start of this section show that the intermediate graph problems used are also
complete for PNP .
Corollary 4.13. Feedback Arc Set Member and Vertex Cover Member are complete for PNP .
Acknowledgements
We thank Lane Hemaspaandra for helpful conversations. We thank two anonymous referees for their helpful
comments.
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